It is currently 20 Jan 2018, 22:37

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

The standard equation of a circle is (x−h)^2+(y−k)^2=r^2, where the

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 43334

Kudos [?]: 139624 [0], given: 12794

The standard equation of a circle is (x−h)^2+(y−k)^2=r^2, where the [#permalink]

Show Tags

New post 29 Nov 2015, 09:51
Expert's post
14
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

69% (01:24) correct 31% (01:52) wrong based on 191 sessions

HideShow timer Statistics

The standard equation of a circle is \((x−h)^2+(y−k)^2=r^2\), where the point (h,k) is the center of the circle and r represents the length of the radius. If a particular circle can be represented by the equation \(x^2+y^2−8x+2y=−1\), what is the radius of that circle?

A. -1
B. 1
C. 2
D. 4
E. 8
[Reveal] Spoiler: OA

_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Kudos [?]: 139624 [0], given: 12794

3 KUDOS received
SC Moderator
User avatar
P
Joined: 13 Apr 2015
Posts: 1540

Kudos [?]: 1294 [3], given: 913

Location: India
Concentration: Strategy, General Management
WE: Information Technology (Consulting)
GMAT ToolKit User Premium Member CAT Tests
Re: The standard equation of a circle is (x−h)^2+(y−k)^2=r^2, where the [#permalink]

Show Tags

New post 29 Nov 2015, 10:23
3
This post received
KUDOS
x^2 + y^2 - 8x + 2y = -1
To bring the above equation into its standard form, complete the squares by adding 16 and 1 on both the sides
x^2 - 8x + 16 + y^2 + 2y + 1 = -1 + 16 + 1
(x - 4)^2 + (y + 1)^2 = 16 --> This is in the form of (x-h)^2 + (y-k)^2 = r^2
r = 4

Answer: D

Kudos [?]: 1294 [3], given: 913

Intern
Intern
avatar
Joined: 14 Mar 2014
Posts: 3

Kudos [?]: [0], given: 2

GMAT ToolKit User Reviews Badge
Re: The standard equation of a circle is (x−h)^2+(y−k)^2=r^2, where the [#permalink]

Show Tags

New post 03 Feb 2016, 17:14
Hi Bunuel,

This is my first time participating in gmatclub. Firs of all thank you for always posting stuff that really helps and for always giving a good solution to math problems.

I've been struggling trying to find out an easy procedure for this question. Can you help?

Thanks again

Juan

Kudos [?]: [0], given: 2

1 KUDOS received
Current Student
avatar
S
Joined: 20 Mar 2014
Posts: 2685

Kudos [?]: 1848 [1], given: 800

Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
GMAT ToolKit User Premium Member Reviews Badge
Re: The standard equation of a circle is (x−h)^2+(y−k)^2=r^2, where the [#permalink]

Show Tags

New post 03 Feb 2016, 18:08
1
This post received
KUDOS
betancurj wrote:
Hi Bunuel,

This is my first time participating in gmatclub. Firs of all thank you for always posting stuff that really helps and for always giving a good solution to math problems.

I've been struggling trying to find out an easy procedure for this question. Can you help?

Thanks again

Juan


Let me try to answer.

You are given that the standard equation of a circle ---> (x−h)^2+(y−k)^2=r^2 where, r=radius of the circle.

Now, you are given another circle with the equation ---> x^2 + y^2 - 8x + 2y = -1 , when you compare this to the standard equation of the circle, you should realize that it would be straightforward to calculate the radius of the circle if you can express x^2 + y^2 - 8x + 2y = -1 in the form of (x−h)^2+(y−k)^2=r^2

In order to make complete squares similar to (x-h)^2 or (y-k)^2, you need to remember that (x+a)^2=x^2+a^2+2ax and (x-a)^2=x^2+a^2-2ax

With these relations in mind, lets go back to x^2 + y^2 - 8x + 2y = -1 with a focus on converting x^2 + y^2 - 8x + 2y to 2 complete squares (1 with x and 1 with y).

x^2 + y^2 - 8x + 2y ---> (x^2-8x+16-16)+(y^2+2y+1-1) = (x-4)^2 -16 + (y+1)^2-1 ---> x^2 + y^2 - 8x + 2y = -1 can thus be written as :

(x-4)^2 -16 + (y+1)^2-1 = -1 --->(x-4)^2 + (y+1)^2 = 16 = 4^2.

NOw you can compare this form to (x−h)^2+(y−k)^2=r^2 to get h = 4, k=-1 and r=radius = 4.

Thus, D is the correct answer.

For questions such as these you need to move backwards from what has been given to you and to modify the given values into a more usable form.

Hope this helps.

Kudos [?]: 1848 [1], given: 800

SVP
SVP
User avatar
B
Joined: 06 Nov 2014
Posts: 1904

Kudos [?]: 564 [0], given: 23

The standard equation of a circle is (x−h)^2+(y−k)^2=r^2, where the [#permalink]

Show Tags

New post 03 Feb 2016, 20:28
1
This post was
BOOKMARKED
betancurj wrote:
Hi Bunuel,

This is my first time participating in gmatclub. Firs of all thank you for always posting stuff that really helps and for always giving a good solution to math problems.

I've been struggling trying to find out an easy procedure for this question. Can you help?

Thanks again

Juan


Hi Juan,

The trick here is to convert everything in the standard form of equation of a circle, which is:
\((x−h)^2+(y−k)^2=r^2\)

We are given the equation of a circle in some other form (expanded form of the standard equation)

\(x^2+y^2−8x+2y=−1\)
On grouping terms,

\(x^2−8x+y^2+2y=−1\)
From here on, we need to make two perfect squares in x and y

The best way to understand which perfect square to make is by dissecting the coefficient of x and y
8 = 2*1*4, 2 = 2*1*1

We are writing the coefficients in this form, because the coefficient of x in \({(x+a)}^2 = x^2 + 2ax + a^2\) (Eq i) is 2*1*a
\(x^2−2*x*4 + y^2+2*1*y = -1\)

Now we need the constant terms to make these perfect squares.
The constant term in Eq (i) is a

Hence we can write \(x^2−2*1*4x + y^2+2*1*y = -1\) as
\(x^2−2*x*4 + 16 + y^2+2*1*y + 1 = -1 + 16 + 1\) (adding constants on both sides of the equation)
\({(x-4)}^2 + {(y+1)}^2 = 16\)
\({(x-4)}^2 + {(y+1)}^2 = 4^2\)

Hence the radius = 4. Option D

Kudos [?]: 564 [0], given: 23

Intern
Intern
avatar
Joined: 14 Mar 2014
Posts: 3

Kudos [?]: [0], given: 2

GMAT ToolKit User Reviews Badge
Re: The standard equation of a circle is (x−h)^2+(y−k)^2=r^2, where the [#permalink]

Show Tags

New post 08 Feb 2016, 15:52
Thank you both, great help.

Kudos [?]: [0], given: 2

Director
Director
avatar
S
Joined: 12 Nov 2016
Posts: 791

Kudos [?]: 39 [0], given: 167

Re: The standard equation of a circle is (x−h)^2+(y−k)^2=r^2, where the [#permalink]

Show Tags

New post 15 Jun 2017, 23:35
Bunuel wrote:
The standard equation of a circle is \((x−h)^2+(y−k)^2=r^2\), where the point (h,k) is the center of the circle and r represents the length of the radius. If a particular circle can be represented by the equation \(x^2+y^2−8x+2y=−1\), what is the radius of that circle?

A. -1
B. 1
C. 2
D. 4
E. 8


Bunuel -

I simply factored (x-h)^2 + (y-k)^2 = r^2

[x^2-2xh + h] + [y^2-2yk + y^2] = r^2

If it can be reduced to

x^2 +y^2 -8x +2y -1

Then don't H and K have to be 4 and -1? Unless I'm making some assumption here that wouldn't necessarily apply to some other example- please let me know

If (x-4)^2 + (y - (-1))^2 = r^2
(4-4)^2 + (1 +1)^2 = r^2
(2)^2= r^2
4= r^2
4 = \sqrt{r}
4= r

Kudos [?]: 39 [0], given: 167

Director
Director
avatar
S
Joined: 12 Nov 2016
Posts: 791

Kudos [?]: 39 [0], given: 167

Re: The standard equation of a circle is (x−h)^2+(y−k)^2=r^2, where the [#permalink]

Show Tags

New post 15 Jun 2017, 23:44
OptimusPrepJanielle wrote:
betancurj wrote:
Hi Bunuel,

This is my first time participating in gmatclub. Firs of all thank you for always posting stuff that really helps and for always giving a good solution to math problems.

I've been struggling trying to find out an easy procedure for this question. Can you help?

Thanks again

Juan


Hi Juan,

The trick here is to convert everything in the standard form of equation of a circle, which is:
\((x−h)^2+(y−k)^2=r^2\)


We are given the equation of a circle in some other form (expanded form of the standard equation)

\(x^2+y^2−8x+2y=−1\)
On grouping terms,

\(x^2−8x+y^2+2y=−1\)
From here on, we need to make two perfect squares in x and y

The best way to understand which perfect square to make is by dissecting the coefficient of x and y
8 = 2*1*4, 2 = 2*1*1

We are writing the coefficients in this form, because the coefficient of x in \({(x+a)}^2 = x^2 + 2ax + a^2\) (Eq i) is 2*1*a
\(x^2−2*x*4 + y^2+2*1*y = -1\)

Now we need the constant terms to make these perfect squares.
The constant term in Eq (i) is a

Hence we can write \(x^2−2*1*4x + y^2+2*1*y = -1\) as
\(x^2−2*x*4 + 16 + y^2+2*1*y + 1 = -1 + 16 + 1\) (adding constants on both sides of the equation)
\({(x-4)}^2 + {(y+1)}^2 = 16\)
\({(x-4)}^2 + {(y+1)}^2 = 4^2\)

Hence the radius = 4. Option D



Yes but don't forget to explain you derived 16 and 1 by taking half of the coefficients 8 and 1 and then squaring them. For example, if we have an algebraic expression that cannot be factored such as x-10x-18=0 - then we complete the square. How?

x-10x __ = 18 (add 18 to both sides or really just put 18 on the other side)
x-10x +25 = 18 + 25 (remember how to get the coefficients? In this equation a= 1 b= 10; take (b/2)^2 and add to both sides
(x-5)(x-5) = 43
x-5 = + or minus 43

Kudos [?]: 39 [0], given: 167

Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 43334

Kudos [?]: 139624 [0], given: 12794

Re: The standard equation of a circle is (x−h)^2+(y−k)^2=r^2, where the [#permalink]

Show Tags

New post 01 Jan 2018, 10:54
Bunuel wrote:
The standard equation of a circle is \((x−h)^2+(y−k)^2=r^2\), where the point (h,k) is the center of the circle and r represents the length of the radius. If a particular circle can be represented by the equation \(x^2+y^2−8x+2y=−1\), what is the radius of that circle?

A. -1
B. 1
C. 2
D. 4
E. 8


VERITAS PREP OFFICIAL SOLUTION:

What should first jump out is that the equation for this particular circle does not match the standard equation of a circle. So you'll need to do some algebra to get to that point.

Your first step should be to factor the x and y terms so that they fit the standard form. Since each term needs to be factored into a squared parenthetical, you should use the common quadratic form \((a−b)^2=a^2−2ab+b^2\) as your guide. Look at the coefficients −8x and +2y, recognizing that they need to fill the −2ab role in that common quadratic equation. This means that you'll factor the x and y terms into:

\((x−4)^2=x^2−8x+16\)
and
\((y−(−1)^2)=y^2+2y+1\)

So now you know that the left-hand side of the standard equation of a circle produces the combination of \(x^2−8x+16+y^2+2y+1\), which equals \(x^2−8x+y^2+2y+17\). So now you know two equations to be true:

\(x^2−8x+y^2+2y+17=r^2\)

and

\(x^2+y^2−8x+2y=−1\)

That second equation can be quickly made to look like the first by adding 1 to both sides:

\(x^2+y^2−8x+2y+1=0\)

Which means that the difference between the two equations is that the left-hand side is 16 apart and the right-hand side is r^2 apart. So to balance, add 16 to the left and r^2 to the right, meaning that r^2=16. This means that the radius is 4, making answer choice D correct.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Kudos [?]: 139624 [0], given: 12794

Re: The standard equation of a circle is (x−h)^2+(y−k)^2=r^2, where the   [#permalink] 01 Jan 2018, 10:54
Display posts from previous: Sort by

The standard equation of a circle is (x−h)^2+(y−k)^2=r^2, where the

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.